Further Mathematics Questions and Answers

Using the remainder theorem, if (x - a) is a factor of f(x), then f(a) = 0.

Check the options and get the answer.

$P = \begin{pmatrix} -2 & 1 \\ 3 & 4 \end{pmatrix}; Q = \begin{pmatrix} 5 & -3 \\ 2 & -1 \end{pmatrix}$

= $PQ = \begin{pmatrix} -10+2 & 6-1 \\ 15+8 & -9-4 \end{pmatrix}$

= $\begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix}$

$QP = \begin{pmatrix} -10-9 & 5-12 \\ -4-3 & 2-4 \end{pmatrix}$

= $\begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}$ 

$PQ - QP = \begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix} - \begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}$

= $\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}$

$\frac{2x}{(x + 6)(x + 3)} = \frac{P}{x + 6} + \frac{Q}{x + 3}$

$\frac{2x}{(x + 6)(x + 3)} = \frac{P(x + 3) + Q(x + 6)}{(x + 6)(x + 3)}$

Comparing equations, we have

$2x = Px + 3P + Qx + 6Q$

$\implies 3P + 6Q = 0 ... (1) ; P + Q = 2 .... (2)$

From equation (1), $3P = -6Q  \implies P = -2Q$

$\therefore -2Q + Q = -Q = 2$

$Q = -2$

$P = -2Q = -2(-2) = 4$

$P = 4, Q = -2$

$f : x \to x^{2} ; g : x \to x + 3$

$g(2) = 2 + 3 = 5$

$f o g(2) = f(5) = 5^{2} = 25$

$(\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...$

$\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}$

= $\frac{8!}{6!2!}(\frac{x^{6}}{64})(1)$

= $28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}$